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If $A=\left|\begin{array}{lll}x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x\end{array}\right|$ and $B=\left|\begin{array}{cc}x & 1 \\ 1 & x\end{array}\right|$, then $\frac{d A}{d x}$ is equal to
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The correct answer is:
$3 B$
$\begin{array}{ll}
\text { Given, } & A=\left|\begin{array}{lll}
x & 1 & 1 \\
1 & x & 1 \\
1 & 1 & x
\end{array}\right| \\
\Rightarrow & A=x\left(x^{2}-1\right)-(x-1)+(1-x) \\
\Rightarrow & A=x^{3}-x-x+1+1-x \\
\Rightarrow & A=x^{3}-3 x+2
\end{array}$
On differentiating w.r.t. $x$, we get
$\frac{d A}{d x}=3 x^{2}-3..(i)$
Also, given $\quad B=\left|\begin{array}{ll}x & 1 \\ 1 & x\end{array}\right|$
$\begin{array}{cc}
\Rightarrow & B=x^{2}-1 \\
\Rightarrow & 3 B=3 x^{2}-3...(ii)
\end{array}$
From Eqs. (i) and (ii), we get
$\frac{d A}{d x}=3 B$
\text { Given, } & A=\left|\begin{array}{lll}
x & 1 & 1 \\
1 & x & 1 \\
1 & 1 & x
\end{array}\right| \\
\Rightarrow & A=x\left(x^{2}-1\right)-(x-1)+(1-x) \\
\Rightarrow & A=x^{3}-x-x+1+1-x \\
\Rightarrow & A=x^{3}-3 x+2
\end{array}$
On differentiating w.r.t. $x$, we get
$\frac{d A}{d x}=3 x^{2}-3..(i)$
Also, given $\quad B=\left|\begin{array}{ll}x & 1 \\ 1 & x\end{array}\right|$
$\begin{array}{cc}
\Rightarrow & B=x^{2}-1 \\
\Rightarrow & 3 B=3 x^{2}-3...(ii)
\end{array}$
From Eqs. (i) and (ii), we get
$\frac{d A}{d x}=3 B$
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